Cos 90=0 Fairly basic
Cos 90 * Cos 90= Cos^2 90 thus 0*0=0
So, Cos^3 90=Cos^2 90*Cos 90=0
and Cos^4 90=Cos^2 90*Cos^2 90=0
0^x=0
Maths Homework
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Cos 90=0 Fairly basic
Cos 90 * Cos 90= Cos^2 90 thus 0*0=0
So, Cos^3 90=Cos^2 90*Cos 90=0
and Cos^4 90=Cos^2 90*Cos^2 90=0
0^x=0
It can get to be incredibly complex (incredible=not credible, I know EXACTLY what I'm saying here and mean it) and yet is composed of the simplest of steps. The impossible is broken down into such simplicity that s/he to whom it is being explained feels like a dunce. (Me, whenever I ask a question in that class) Certainty goes much further than the certainty of other fields. With other stuff it is a "close enough, it's proven" attitude, whereas with math, if something is proven, it is irrefutable. Such an inexplicable subject, I'd rather do as little with it as possible.
Originally posted by Bot_40
OK claw, now you've got your chance to prove yourself
Find all values of x when 0<x<360 for:
9sin²x - 6sin x + cos² x = 0
I think that's the only other one I can't do out of the 35. Then I'm done with trig. identities for good
That is ... until the exams
Ok, I’ll help you a little, but this won't do you any good unless you start making diagrams and thinking for yourself. You're going to get a test and not know what to do.
The only values you will be concerned with are ones and zeros. Here are your values at x = 0, pi/2, pi, 3pi/2, and 2pi:
Sine: 0 1 0 –1 0
Cosine: 1 0 –1 0 1
You’ll see that the only difference between sine and cosine is that one is just shifted 90 degrees relative to the other.
Anyway, you have an “equation.” This means that you have to have exactly the same thing on both sides of the equal sign. I know you’re probably going, “Duh, like I didn’t know that.” but it’s an important point here. What values can you plug in that would make the left side of the equation 0 so it matches the 0 on the right side?
Well, if you think about it, any values for sine that are either one or negative one is going to mean you’ll have nine times one (or negative one) or six times one (or negative one). This would give you 9 – 6, which is 3, or some integer that is going to be higher than one. This is no good. The only hope you have is for the sine values to be zero and the cosine value to be zero. Is it possible for the sine and cosine to be zero at the same time so you have: 9*0 – 6*0 + 0^2 = 0? You tell me? If you graph the sine and cosine functions, you’ll see that one always lags behind the other by 90 degrees/pi/2, which means they will never both be 0 at the same time.
So, is there a solution, or do you write: No solution?
Originally posted by Bot_40
OK claw, now you've got your chance to prove yourself
Find all values of x when 0<x<360 for:
9sin²x - 6sin x + cos² x = 0
I think that's the only other one I can't do out of the 35. Then I'm done with trig. identities for good
That is ... until the exams
X = 14.477512186
X = 30
X = 150
X = 165.52248781
Hahaha haha You’re not helping.
I’m trying to get him to reason things out, question things, and ask himself how should I approach the problem. He’s going to get on a test and not know what to do. Unless his math classes are different from mine, you’re not allowed to use a TI-83 on the test. We had to do everything out by hand. Not that it was any problems, because if you knew what you were doing and worked the problems right, the numbers were easy.Your answers with the decimals are not an “exact” answer. They don’t do things like that in trig. They want an exact answer, unless, again, your classes are different from mine. You have to convert sines and cosines, and then do the problem. You do know how to convert them into something else besides sines and cosines, right? You can convert this stuff into algebraic equations and work the problems and get an “exact” answer. You are doing trig identities, which means one thing can be identified in more than one way.
